3.482 \(\int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=239 \[ \frac{2 a (99 A+88 B+80 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{4 (99 A+88 B+80 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac{8 (99 A+88 B+80 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{11 d} \]

[Out]

(4*a*(99*A + 88*B + 80*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(99*A + 88*B + 80*C)*Sec[c + d
*x]^3*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(11*B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt
[a + a*Sec[c + d*x]]) - (8*(99*A + 88*B + 80*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*C*Sec[c +
d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (4*(99*A + 88*B + 80*C)*(a + a*Sec[c + d*x])^(3/2)*Tan
[c + d*x])/(1155*a*d)

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Rubi [A]  time = 0.55846, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4088, 4016, 3803, 3800, 4001, 3792} \[ \frac{2 a (99 A+88 B+80 C) \tan (c+d x) \sec ^3(c+d x)}{693 d \sqrt{a \sec (c+d x)+a}}+\frac{4 (99 A+88 B+80 C) \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{1155 a d}-\frac{8 (99 A+88 B+80 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{3465 d}+\frac{4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt{a \sec (c+d x)+a}}+\frac{2 a (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{99 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^4(c+d x) \sqrt{a \sec (c+d x)+a}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(4*a*(99*A + 88*B + 80*C)*Tan[c + d*x])/(495*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(99*A + 88*B + 80*C)*Sec[c + d
*x]^3*Tan[c + d*x])/(693*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*(11*B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(99*d*Sqrt
[a + a*Sec[c + d*x]]) - (8*(99*A + 88*B + 80*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3465*d) + (2*C*Sec[c +
d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + (4*(99*A + 88*B + 80*C)*(a + a*Sec[c + d*x])^(3/2)*Tan
[c + d*x])/(1155*a*d)

Rule 4088

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]

Rule 4016

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] &&  !LtQ[n, 0]

Rule 3803

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*d
*Cot[e + f*x]*(d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(2*a*d*(n - 1))/(b*(
2*n - 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*Sqrt[a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{2 \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \left (\frac{1}{2} a (11 A+8 C)+\frac{1}{2} a (11 B+C) \sec (c+d x)\right ) \, dx}{11 a}\\ &=\frac{2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{99} (99 A+88 B+80 C) \int \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{1}{231} (2 (99 A+88 B+80 C)) \int \sec ^3(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{4 (99 A+88 B+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac{(4 (99 A+88 B+80 C)) \int \sec (c+d x) \left (\frac{3 a}{2}-a \sec (c+d x)\right ) \sqrt{a+a \sec (c+d x)} \, dx}{1155 a}\\ &=\frac{2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}-\frac{8 (99 A+88 B+80 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{4 (99 A+88 B+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}+\frac{1}{495} (2 (99 A+88 B+80 C)) \int \sec (c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt{a+a \sec (c+d x)}}+\frac{2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt{a+a \sec (c+d x)}}-\frac{8 (99 A+88 B+80 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac{2 C \sec ^4(c+d x) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac{4 (99 A+88 B+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d}\\ \end{align*}

Mathematica [A]  time = 1.72971, size = 185, normalized size = 0.77 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt{a (\sec (c+d x)+1)} ((2871 A+3322 B+3020 C) \cos (c+d x)+13 (99 A+88 B+80 C) \cos (2 (c+d x))+1287 A \cos (3 (c+d x))+198 A \cos (4 (c+d x))+198 A \cos (5 (c+d x))+1089 A+1144 B \cos (3 (c+d x))+176 B \cos (4 (c+d x))+176 B \cos (5 (c+d x))+968 B+1040 C \cos (3 (c+d x))+160 C \cos (4 (c+d x))+160 C \cos (5 (c+d x))+1510 C)}{3465 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((1089*A + 968*B + 1510*C + (2871*A + 3322*B + 3020*C)*Cos[c + d*x] + 13*(99*A + 88*B + 80*C)*Cos[2*(c + d*x)]
+ 1287*A*Cos[3*(c + d*x)] + 1144*B*Cos[3*(c + d*x)] + 1040*C*Cos[3*(c + d*x)] + 198*A*Cos[4*(c + d*x)] + 176*
B*Cos[4*(c + d*x)] + 160*C*Cos[4*(c + d*x)] + 198*A*Cos[5*(c + d*x)] + 176*B*Cos[5*(c + d*x)] + 160*C*Cos[5*(c
+ d*x)])*Sec[c + d*x]^5*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(3465*d)

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Maple [A]  time = 0.433, size = 204, normalized size = 0.9 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( 1584\,A \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1408\,B \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1280\,C \left ( \cos \left ( dx+c \right ) \right ) ^{5}+792\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}+704\,B \left ( \cos \left ( dx+c \right ) \right ) ^{4}+640\,C \left ( \cos \left ( dx+c \right ) \right ) ^{4}+594\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}+528\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+480\,C \left ( \cos \left ( dx+c \right ) \right ) ^{3}+495\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}+440\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+400\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}+385\,B\cos \left ( dx+c \right ) +350\,C\cos \left ( dx+c \right ) +315\,C \right ) }{3465\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

-2/3465/d*(-1+cos(d*x+c))*(1584*A*cos(d*x+c)^5+1408*B*cos(d*x+c)^5+1280*C*cos(d*x+c)^5+792*A*cos(d*x+c)^4+704*
B*cos(d*x+c)^4+640*C*cos(d*x+c)^4+594*A*cos(d*x+c)^3+528*B*cos(d*x+c)^3+480*C*cos(d*x+c)^3+495*A*cos(d*x+c)^2+
440*B*cos(d*x+c)^2+400*C*cos(d*x+c)^2+385*B*cos(d*x+c)+350*C*cos(d*x+c)+315*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(
1/2)/cos(d*x+c)^5/sin(d*x+c)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.517519, size = 405, normalized size = 1.69 \begin{align*} \frac{2 \,{\left (16 \,{\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{5} + 8 \,{\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \,{\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{2} + 35 \,{\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right ) + 315 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/3465*(16*(99*A + 88*B + 80*C)*cos(d*x + c)^5 + 8*(99*A + 88*B + 80*C)*cos(d*x + c)^4 + 6*(99*A + 88*B + 80*C
)*cos(d*x + c)^3 + 5*(99*A + 88*B + 80*C)*cos(d*x + c)^2 + 35*(11*B + 10*C)*cos(d*x + c) + 315*C)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 4.98062, size = 554, normalized size = 2.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2/3465*(3465*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 3465*sqrt(2)*C*a^6*sgn(
cos(d*x + c)) - (10395*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 8085*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 5775*sqrt(2)*C
*a^6*sgn(cos(d*x + c)) - (15246*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 14322*sqrt(2)*B*a^6*sgn(cos(d*x + c)) + 1617
0*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (14058*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 13266*sqrt(2)*B*a^6*sgn(cos(d*x +
c)) + 8910*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (6633*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 4741*sqrt(2)*B*a^6*sgn(c
os(d*x + c)) + 5885*sqrt(2)*C*a^6*sgn(cos(d*x + c)) - (891*sqrt(2)*A*a^6*sgn(cos(d*x + c)) + 1177*sqrt(2)*B*a^
6*sgn(cos(d*x + c)) + 755*sqrt(2)*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan
(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/
2*c)^2 - a)^5*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d)